Minimax Nonparametric Parallelism Test

Testing the hypothesis of parallelism is a fundamental statistical problem arising from many applied sciences. In this paper, we develop a nonparametric parallelism test for inferring whether the trends are parallel in treatment and control groups. In particular, the proposed nonparametric parallelism test is a Wald type test based on a smoothing spline ANOVA (SSANOVA) model which can characterize the complex patterns of the data. We derive that the asymptotic null distribution of the test statistic is a Chi-square distribution, unveiling a new version of Wilks phenomenon. Notably, we establish the minimax sharp lower bound of the distinguishable rate for the nonparametric parallelism test by using the information theory, and further prove that the proposed test is minimax optimal. Simulation studies are conducted to investigate the empirical performance of the proposed test. DNA methylation and neuroimaging studies are presented to illustrate potential applications of the test. The software is available at https://github.com/BioAlgs/Parallelism.

• Section S.1 includes some auxiliary lemmas in proving Theorem 8.
• Section S.2 includes the proof of Corollary 9 and Corollary 10.

S.1. Some Auxiliary Lemmas in Proving Theorem 8
We first introduce several notations and lemmas and then start the main proof of Theorem 8. Let g * = f 10 + f 11 and its estimator as Lemma S.1 If f H < 1 for any f ∈ H, as n → ∞, λ → 0 and λ ≥ n −1 , we have where c is a constant, || • || n is the empirical norm.
In the following Lemma S.2, we discuss the relationship between the empirical norm and L 2 norm.Recall the definition of empirical norm and L 2 norm are as follows: Lemma S.2 Under the quasi-uniform design or the uniform design, for f : i.e. the empirical norm of f dominates the L 2 norm.

S.2. Proofs of Corollary 9 and Corollary 10
In order to find the optimal distinguishable rate, we need to bound the tail sum of the eigenvalues of the empirical kernel matrix.We state the following two lemmas which give upper bounds for the tail sum of the eigenvalues of the empirical kernel matrix under the quasi-uniform design and the uniform design respectively.
Lemma S.3 (Liu et al. ( 2019)) If 1/n < λ → 0 and the quasi-uniform design is satisfied, then with probability at least where C > 0 is an absolute constant.
Lemma S.4 If λ > 0 and the uniform design is satisfied, we have where C > 0 is an absolute constant.
Now we start the main proof of Corollary 9 and Corollary 10.The distinguishable rate is where σ 2 n,λ = 2θ 4 11 σ 4 Tr(∆ 2 )/n 2 .We now derive the order of σ 2 n,λ .Since the eigenvalues of ∆ are less than 1, and by Lemma 15, we have Proof Under the quasi-uniform design, applying Lemma A.1, we have with probability at least 1 − 4e Similarly, we have Equation (S.2) satisfied under the uniform design by applying Lemma 14 and Lemma S.4.
Thus we have the minimum distinguishable rate 4m+1) .
By Lemma S.2, this optimal distinguishable rate is achieved in the sense of L 2 norm.

S.3.1. PROOF OF LEMMA 12
Proof Write matrix R as , where θ d = θ 01 − θ 11 .The inverse of M can be written as , and I n denotes the n × n identity matrix.Note that S is a 2n × 2 matrix defined as S = (1 n , 1 n ) T .We thus have where the second equality holds by the fact a − 2b = 0 and Woodbury matrix identity.Note that Therefore, we have ) are symmetric matrices and their diagnonal entries are identical, we have

S.3.2. PROOF OF LEMMA 13
Proof Under the quasi-uniform design, X 1 1 ,. . ., X n n are i.i.d with distribution ω 1 .Therefore, by Theorem 3 in Braun (2006), for 1 ≤ i ≤ n and i ≤ r ≤ n, simple algebra yields where Λ r = ∞ i=r+1 µ i , C is an absolute constant, and c m is a constant depending solely on m.Since the eigenvalue µ i has the polynomial decay rate i −2m , we have . Next, we have, for any i = 1, . . ., n 1 2m − , the empirical eigenvalue μi satisfies where c = c 2 m 2C 4 , c m is a constant only related to m, and M is an absolute constant.To ensure the probability in Equation (S.4) goes to 1, we further require m > 3/2.Thus, for λ > 1/n and m > 3/2, τλ τ λ with probability at least 1 − (n

S.3.3. PROOF OF LEMMA 14
Proof Considering H 1 as the homogeneous Sobolev space, the kernel function K 1 1 can be explicitly written as Under the uniform design, we have the X 1 1 , . . ., X 1 n evenly distributed on [0, 1].Without loss of generality, we assume that X ) which is a symmetric circulant matrix of order n (Shang and Cheng, 2017) with eigenvalues where I n is the n × n identity matrix, and θ d = θ 01 − θ 11 .Letting E = D + 2λI n = Diag{μ i + 2λ} and F = θ d D = Diag{θ d μi }, we have Using the inverse of block matrix, we have where (S.8) Consequently,

We thus have
where τλ is the effective dimension for kernel matrix K 11 .For the any i < τλ , we have μi μi +2λ > 1 3 .
Now we shall prove the upper bound for Tr(∆).Since Tr(∆) has the expression in Equation (S.10), we expand V 11 as and the ith diagonal entry of (E (S.13) Combining Equation (S.12) and Equation (S.13), we have the ith diagonal entry of V 11 Since the lower diagonal block of DE −1 is identical to the upper diagonal block, we only need to bound the trace of DE −1 .We have μi .

Thus we have Tr
S.3.5.PROOF OF LEMMA S.1 Proof By the functional decomposition in Equation ( 10 Combining Equation (S.14) and Equation (S.15), we have By Equation (S.1), we have Noting that M = R + λI n , the eigenvalues of I n − R(R + λI n ) −1 are all smaller than 1, and the rank of where the last inequality holds by applying Woodbury matrix identity, and Equation (S.16).The proof is thus completed.(S.17) Since the population eigenvalues are {(2πi) −2m } ∞ i=1 , we calculate the population efficient dimension as τ λ = (λ) −1/2m /2π.By the inequalities Equation (S.6), we have μ * i ≥ λ for i = 1, . . ., τ λ or i = n − τ λ , . . ., n.We can bound the term in Equation (S.17) By the upper bound of μ * i given in Equation (S.6), we have which completes the proof.